Solved Problems for Chapter 3

If possible, express the vector
\begin{equation*} \vec {v}= \begin{bmatrix} 4 \\ 4 \\ -3 \end{bmatrix} \end{equation*}
as a linear combination of the vectors
\begin{equation*} \vec {u}_1 = \begin{bmatrix} 3 \\ 1 \\ -1 \end{bmatrix} \mbox { and } \vec {u}_2 = \begin{bmatrix} 2 \\ -2\\ 1 \end{bmatrix}. \end{equation*}

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We start with a linear combination equation \(a\begin{bmatrix}3\\1\\-1\end{bmatrix}+b\begin{bmatrix}2\\-2\\1\end{bmatrix}=\begin{bmatrix}4\\4\\-3\end{bmatrix}\). This gives us the following:
\[\left [ \begin{array}{rr|r} 3 & 2 & 4 \\ 1 & -2 & 4\\ -1 & 1 & -3 \end{array} \right ] \rightsquigarrow \left [ \begin{array}{rr|r}1& 0& 2\\ 0& 1 &-1\\ 0& 0& 0\end{array} \right ]\]
This gives us \(\vec {v}=2\vec {u}_1 -\vec {u}_2\).
Decide whether
\[ \vec {v}= \begin{bmatrix} 4 \\ 4 \\ 4 \end{bmatrix} \]
is a linear combination of the vectors
\[ \vec {u}_1 = \begin{bmatrix} 3 \\ 1 \\ -1 \end{bmatrix} \mbox { and } \vec {u}_2 = \begin{bmatrix} 2 \\ -2\\ 1 \end{bmatrix}. \]

Click the arrow to see answer.

\[\left [ \begin{array}{rr|r} 3 & 2 & 4 \\ 1 & -2 & 4\\ -1 & 1 & 4 \end{array} \right ] \rightsquigarrow \left [ \begin{array}{rr|r}1& 0& 0\\ 0& 1 & 0\\ 0& 0& 1\end{array} \right ]\]
The last row indicates that there is no solution to the linear combination equation \(a\vec {u}_1+b\vec {u}_2=\vec {v}\). Therefore, \(\vec {v}\) is not a linear combination of \(\vec {u}_1\) and \(\vec {u}_2\).
Determine whether the following set of vectors is linearly independent.
\[\left \{\begin{bmatrix} 2\\ -3\\0 \end{bmatrix}, \begin{bmatrix} 1\\0\\1 \end{bmatrix}, \begin{bmatrix} 1\\1\\4 \end{bmatrix}\right \}\]

Click the arrow for answer.

The linear combination equation \(a\begin{bmatrix} 2\\ -3\\0 \end{bmatrix}+ b\begin{bmatrix} 1\\0\\1 \end{bmatrix}+ c\begin{bmatrix} 1\\1\\4 \end{bmatrix}=\vec {0}\) has only the trivial solution. We can see this as follows.
\[\text {rref}\left (\begin{bmatrix}2 & 1 & 1\\-3 & 0 & 1\\0 & 1 & 4\end{bmatrix}\right )=\begin{bmatrix} 1 & 0 & 0\\0 & 1 &0\\ 0 & 0 &1 \end{bmatrix}\]
Therefore the given set of vectors is linearly independent.
Determine whether the following set of vectors is linearly independent.
\[\left \{\begin{bmatrix} 2\\ -3\\0 \end{bmatrix}, \begin{bmatrix} 4\\-1\\8 \end{bmatrix}, \begin{bmatrix} 1\\1\\4 \end{bmatrix}\right \}\]

Click the arrow for answer.

The linear combination equation \(a\begin{bmatrix} 2\\ -3\\0 \end{bmatrix}+ b\begin{bmatrix} 4\\-1\\8 \end{bmatrix}+ c\begin{bmatrix} 1\\1\\4 \end{bmatrix}=\vec {0}\) has a non-trivial solution. We can see this as follows.
\[\text {rref}\left (\begin{bmatrix}2 & 4 & 1\\-3 & -1 & 1\\0 & 8 & 4\end{bmatrix}\right )=\begin{bmatrix} 1 & 0 & -1/2\\0 & 1 &1/2\\ 0 & 0 &0 \end{bmatrix}\]
Therefore the given set of vectors is linearly dependent.
Suppose \(\left \{ \vec {x}_{1},\ldots ,\vec {x}_{k}\right \} \) is a set of vectors from \(\RR ^{n}.\) Show that \(\vec {0}\) is in \(\mbox { span}\left \{ \vec {x}_{1},\ldots ,\vec {x}_{k}\right \} .\)

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Let all of the coefficients in the linear combination be zero: \(\sum _{i=1}^{k}0\vec {x}_{k}=\vec {0}\)
True or false?

Click the arrow to see each answer.

1.
The span of any one vector in \(\RR ^n\) is a line.

This is technically false because the span of the zero vector is a point. For all non-zero vectors this statement is true.
2.
The span of two non-zero vectors in \(\RR ^n\) is a plane.

This is false. If two vectors are scalar multiples of each other (i.e. they are linearly dependent), then their span is a line, not a plane.
3.
Given two linearly independent vectors, \(\vec {v}\) and \(\vec {w}\), let \(\vec {n}\) be a normal vector to the plane spanned by \(\vec {v}\) and \(\vec {w}\). The set \(\left \{\vec {v}, \vec {w}, \vec {n}\right \}\) is linearly independent.

This is true. A normal to the plane is a vector perpendicular to the plane. Only vectors that lie in the plane spanned by \(\vec {v}\) and \(\vec {w}\) can be expressed as a linear combination of \(\vec {v}\) and \(\vec {w}\). Since \(\vec {n}\) is not in the plane, it cannot be expressed as a linear combination of \(\vec {v}\) and \(\vec {w}\). Therefore the set \(\left \{\vec {v}, \vec {w}, \vec {n}\right \}\) is linearly independent.

Bibliography

Some of the Review Exercises come from the end of Chapter 4 of Ken Kuttler’s A First Course in Linear Algebra. (CC-BY)

Ken Kuttler, A First Course in Linear Algebra, Lyryx 2017, Open Edition, pp. 151–152.